[extropy-chat] MARS: Because it is hard
dgc at cox.net
Thu Apr 15 04:29:24 UTC 2004
>>Spike: how fast is ELLO? Surely 100KM is long enough to stay below 1G?
>Lets estimate it. The diameter of the moon is about 3500 km,
>the mass is about one eightieth the earth, so about 7.4e22.
>omega^2*R = 7.4E22*6.7E-11/3.5E6^2
>so omega = 3.4E-4 and omega*R is the tangential velocity
>of about 3.4E-4*3500 is about 1200 meters per second.
>One G is about 10 m/s so 1200 = A*t so t = 120 seconds,
>length of the track is .5*10*t^2 = 5*120^2 = 70 km of
>track needed to accelerate at one G.
This differs form Alan's by a factor of two. Radius versus diameter?
If Alan's is correct (I haven't checked) I rounded up to 150Km.
OK, how much energy can we dissipate or recapture using a linear
induction motor system, practically speaking. Assume a 10,000Kg
capsule and 150Km decel from 1200m/s. We must dissipate the
energy of 10,000g every 150m, or 1g every 15mm, about like stopping
a .22cal bullet. This seems to be more like engineering than it is like
magic. I'm assuming that a 10,000Kg capsule would be a convenient
More information about the extropy-chat